202. Happy Number - Detailed Explanation

In this brute-force approach, the key is the use of a HashSet (seen) to store numbers that have appeared in the sequence. If we ever generate a number that we've seen before (i.e., n in seen), it means we are in a loop and will never reach 1, so we can return false at that point. If we exit the loop, it must be because n == 1, in which case we return true.

Approach 2: Floyd’s Cycle Detection (Tortoise and Hare Algorithm)

Idea:
This approach uses the cycle-detection technique known as Floyd’s Tortoise and Hare algorithm (often used to detect cycles in linked lists). The concept is to use two "pointers" that move through the sequence at different speeds:

• Slow pointer (tortoise): Moves one step (one transformation) at a time.

• Fast pointer (hare): Moves two steps at a time (i.e., apply the transformation twice for each step of the fast pointer).

We initialize slow = n and fast = next_number(n) where next_number(x) is the function that returns the sum of the squares of the digits of x. Then we loop:

• Move slow to next_number(slow) (one step).

• Move fast to next_number(next_number(fast)) (two steps in one iteration).

• If at any point fast becomes 1, then the number is happy (because the fast pointer managed to reach 1, which implies the sequence leads to 1). Return true.

• If at any point slow == fast (and it’s not 1), that means the two pointers have met in a cycle. This indicates we are in an endless loop that doesn't include 1, so return false.

The reasoning is that if there is a cycle, the fast pointer will eventually lap the slow pointer and they will meet inside the cycle. If there is no cycle (the sequence ends in 1), the fast pointer will hit 1 and stop before they ever meet at a non-1 number.

Complexity Analysis:

• Time Complexity: O(log n), for the same reasoning as the first approach – each transformation is O(log n) in the number of digits. The number of transformations is again limited because either we find 1 or detect a cycle fairly quickly. The cycle detection does not add extra complexity in big-O terms; it's just a constant amount of work per iteration.

• Space Complexity: O(1), which is an improvement over Approach 1. We do not use extra space proportional to n or the length of the sequence; we only use a few integer variables (slow, fast, etc.). We don't store the entire sequence or seen numbers.

Python Code:

Using Floyd’s algorithm avoids the need for extra memory to track seen numbers. If n is not happy, the two pointers will eventually be trapped in the same cycle and meet each other, at which point we know to return false. If n is happy, the fast pointer will hit 1 and we return true. This is an optimal solution in terms of space.

Step-by-Step Walkthrough of Solution

Let's break down the procedure in simple steps using either approach:

1. Initialize: Start with the given number n. (For the cycle detection approach, also initialize slow = n and fast = n or fast = next_number(n).)

2. Compute Next Number: Define a function or process to compute the sum of the squares of the digits of the current number. For example, if the current number is 256, the next number is 2^2 + 5^2 + 6^2 = 4 + 25 + 36 = 65.

3. Iterate: Update the current number to this next number and repeat the process.

   • In the hash set approach, add each number to a set of seen numbers before moving to the next.
   • In the cycle detection approach, move the slow and fast pointers as described.

4. Check for 1: If at any point the current number (or the fast pointer) becomes 1, stop. The answer is true because we have found that n is a happy number (the sequence ended in 1).

5. Check for Cycles:

   • In the hash set approach, before computing the next number, check if the current number is already in the seen set. If it is, that means we have a cycle, so stop and return false (not a happy number).

   • In the cycle detection approach, continue until slow == fast indicates a cycle (and the number is not 1), then return false.

6. Repeat: Continue steps 2-5 until you either return true or false.

By following these steps, you ensure that you will catch the two possible end conditions: reaching 1 (happy) or repeating a number (cycle, not happy). You won't loop forever because one of those conditions will eventually occur for any positive integer n.

Common Mistakes and Pitfalls

When implementing the solution for the happy number problem, watch out for these common mistakes:

1. Not handling cycles (infinite loops): If you simply keep calculating the next number without any cycle detection, you might run forever for an unhappy number. Always include a mechanism (like a set or the two-pointer method) to break out of cycles.

2. Not checking for 1 early: It's efficient to check if the number is 1 before or immediately after computing the next number. If you delay this check, you might do unnecessary computations. Also, if the input n is 1 to begin with, you should return true immediately without entering any loop.

3. Ignoring integer overflow (in some languages): In Python, integer overflow is not a concern because Python integers have arbitrary precision. However, in languages like Java, although the sum of squares of digits of a 32-bit integer won't overflow a 32-bit range (the maximum sum of squares for a 10-digit number like 2,147,483,647 is 9^2 * 10 = 810, which is well within int range), it's still good practice to ensure you use an appropriate data type. In our Java solution, we stayed within int safely. Just be mindful of numeric types in other contexts.

Edge Cases to Consider

• n = 1: This is the smallest happy number. The process is trivial (it is already 1, so return true).

• n = 10: This number is happy, and it's a simple case because 1^2 + 0^2 = 1 in one step. It’s good to verify that the code handles numbers with a 0 correctly in their digits.

• Large n (near the upper bound): Even for a large number like 2^{31}-1 = 2147483647, the process will quickly reduce the number. It's useful to test a large unhappy number to ensure your cycle detection works and you don't run into performance issues.

• Single-digit numbers: Aside from 1, other single-digit numbers will be happy if and only if they are 1, 7 (since 7 goes to 49 → ... → 1 as shown), or maybe others? Actually, 7 is the only single-digit >1 that is happy. All other single digits (2-9 except 7) eventually lead to an unhappy cycle. It's good to test those basics.

Alternative Variations and Extensions

This problem can be extended or varied in a few interesting ways:

1. Find the cycle length for unhappy numbers: Instead of just detecting a cycle, you could determine how long the cycle is (for example, the cycle 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4 has length 8).

2. Different powers or bases: The classic happy number uses squares (power of 2) in base 10. You could define a similar process using cubes of digits (power of 3) or in different number bases, and ask if a number is "happy" under those rules. (In general, in number theory, there's a concept of b-happy numbers using the sum of pth powers of digits.

3. Next happy number: Given an input n, find the smallest happy number greater than n. This would involve checking numbers one by one above n until you find one that is happy.

Related Problems for Further Practice

• Linked List Cycle Detection: (Floyd's Tortoise and Hare algorithm) This is the classic application of cycle detection in a linked list, which inspired Approach 2. It’s good practice for understanding cycle detection in general.

• Power of Two: A number theory problem where you determine if a number is a power of two. This is simpler but also involves recognizing patterns in binary or through mathematical properties.

• Digital Root (Add Digits): Repeatedly sum the digits of a number until you get a single digit. This is a different iterative digit process with a known formula, but conceptually similar in that it repeatedly transforms a number based on its digits.

• Reverse Integer: Although unrelated to happy numbers, it's another common problem that involves manipulating digits of an integer (reversing them) and dealing with edge cases like overflow or negative numbers.